How many liters of pure alcohol must be added to a 100-liter

Question

How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol?

(A) 7/2
(B) 5
(C) 20/3
(D) 8
(E) 39/4

fluke · Accepted Answer

20% Alcohol solution means; in the 100 liter solution, 20 liters of solution is alcohol and 80 liters other solvents.

If we add "x" liters of alcohol to the solution, the solution becomes "100+x" liters and alcohol, which was 20 liters, becomes 20+x liters.

According to the statement; 
20+x = 25% of (100+x)
OR
20+x=(100+x)/4
80+4x=100+x
3x=20
x=20/3

Ans: "C"

PareshGmat · Answer

Alcohol .......... Non-Alcohol .................... Total

20 ..................... 80 ................................. 100

20+x ................... 80 .............................. 100+x

Setting up the equation

\frac{75}{100} (100+x) = 80  OR

\frac{25}{100}(100+x) = 20+x

x = \frac{20}{3}

Answer = C

plb · Answer

Here is my crack at it:

Given :
 
 20% of solution is alcohol 
 80% of solution is other 
 Total volume = 100 liters

Derived :
 
 20% * 100 liters = 20 liters

Now, if we add additional alcohol to the solution, we are altering the volume, so it's no longer going to be just 100 liters, it's going to be 100 plus some arbitrary amount. This gives us 100 + x liters of solution after adding more alcohol. In addition, when we added more alcohol, there is not just 20 liters of alcohol anymore (we just added some more), there is going to be 20 + x liters of alcohol.

What we're wanting to figure out is: how much alcohol must be added to result in there being 25% alcohol.

As mentioned before, we started out with:

20% * 100 = 20 liters

But, now we have:

25% * (100 + x) = (20 + x)

Now, let's just solve for x:

Convert the percentage and multiply:
 {
\frac{25}{100} * (100 + x) = (20 + x)
}

{
\frac{(2500 + 25x)}{100} = (20 + x)
}

Simply the left side:
 {
\frac{25(100 + x)}{100} = (20 + x)
}

Cancel out terms:
 {
\frac{(100 + x)}{4} = (20 + x)
}

Multiple both sides by 4 to get rid of the 4 in the denominator on the left:
 {
4 \bigg( \frac{(100 + x)}{4} \bigg) = 4(20 + x)
}

{
(100 + x) = (80 + 4x)
}

Subtract the x from the left:
 {
(100) = (80 + 3x)
}

Subtract the 80 from the right:
 {
20 = 3x
}

Divide....success!
 {
x = \frac{20}{3}
}

subhashghosh · Answer

Equating total volume of alcohol :

(x + 100) * 25/100 = x + 20

=> x/4 + 25 = x + 20

=> 3x/4 = 5

=> x = 20/3

Answer - C

prj100 · Answer

Using Alligations:

We have 20% alcohol and 80% water. As per the question in the final mixture alcohol should be 25% and water should be 75%. Hence we can say that 80% water will be mixed with 0% water to yield a solution that will contain 75% water. So using alligations:

80%---------0%

75

75-----------5
75:5 = 15:1

Say we added x units of alcohol to the original mixture. These x units will have 0% water. And from above we see 80% water solution is mixed with 0% water solution in 15:1.

so, 15/1 = 100/x
x= 100/15
x=20/3 is the answer

yenh · Answer

We can solve such weighted average problem by understanding the underlying concept. You need to look at the difference between the alcohol level of the 20% solution and the pure alcohol.
_ The 20% solution has a alcohol level that is 5% lower than the level of alcohol of the final mixture, let's say -5 differential.
_ The pure (100%) alcohol has a alcohol level that is 75% higher than the level of alcohol of the final mixture, let's say +75 differential.
You need to make these differentials cancel out, so just multiply both differentials by respective volumes so that the positive will cancel out with the negative: 75x - 5.100 = 0 -> x = 500/75 = 20/3

Hope it's clear.

EMPOWERgmatRichC · Answer

Hi All,

Since this question is built around the concept of a "weighted average", there are several different ways that you can go about solving it. Since the question asks how much alcohol must be added, and the answer choices are NUMBERS, we can TEST THE ANSWERS and do some basic arithmetic to get to the solution.

We're told that a 100-liter solution is 20% alcohol. This means that 20 liters are alcohol and 80 liters are not alcohol. We're asked how many liters must be added to this 100-liter solution to get a solution that is 25% alcohol.

Let's start with ANSWER B: 5 liters
IF....we add 5 liters of alcohol to this existing mixture....
Total alcohol = 20+5 = 25 liters
Total liquid = 100+5 = 105 liters
25/105 is LESS than 25% alcohol (since 25/100 = 25%, 25/105 is LESS than 25%), thus Answer B CANNOT be the answer. We need MORE alcohol to raise the percentage.
Eliminate Answers A and B.

Next, let's TEST ANSWER D: 8 liters
IF....we add 8 liters of alcohol to this existing mixture....
Total alcohol = 20+8 = 28 liters
Total liquid = 100+8 = 108 liters
28/108 is MORE than 25% alcohol (since 28/112 = 25%, 28/108 is MORE than 25%), thus Answer D CANNOT be the answer. We need LESS alcohol to raise the percentage.
Eliminate Answers D and E.

There's only one answer left...

Final Answer:  C

GMAT assassins aren't born, they're made,
Rich

Abhishek009 · Answer

\frac{Alcohol}{Solution} = \frac{20+x}{100+x} = \frac{1}{4}

\frac{Alcohol}{Solution} => 80 + 4x = 100 + x

\frac{Alcohol}{Solution} => 3x = 20

Or,  x = \frac{20}{3}

Hence, answer must be (C)  \frac{20}{3}

KarishmaB · Answer

Responding to a pm:

Using scale method here:

Working with the concentration of alcohol,

w1/w2 = (A2 - Aavg)/(Aavg - A1)

w1/w2 = (100 - 25)/(25 - 20) = 75/5 = 15/1

So for every 15 parts of the 20% solution, we must put 1 part of pure alcohol.

For 100 liter solution, we need 100/15 = 20/3 liters of pure alcohol.

AkshdeepS · Answer

Mam, why did we use A2 as 100. It is in litres and Aavg and A1 as percentage. I was doing A2 = 25, Aavg = 20, A1 = 0 as we are using pure alchohol.

KarishmaB · Answer

We are working with the percentages of alcohol. 20% alcohol solution. 25% alcohol solution. Pure alcohol has 100% alcohol.

You would use 0 if you were working with the percentages of water.

KarishmaB · Answer

w1 and w2 are weights of solutions 1 and 2. They are not amount of alcohol in each solution. They are the weights of entire solutions.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

How many liters of pure alcohol must be added to a 100-liter

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

How many liters of pure alcohol must be added to a 100-liter

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards